Electrical – Converting full adder to subtractor using inverter

The standard adder / subtract circuit is given by the figure below.For an explanation of how it works please see my previous answer here : Subtraction using adder circuit

Now notice that if you were to change the input that is tied to '0' to Cin, then the circuit would now perform the operation: A + ~B + (Cin ^ SUB), meaning that:

1) it would be a normal Adder/Subtract circuit when Cin is 0.

2) if Cin = 1 and Sub = 1 it would evaluate to:

A + ~B + (1 ^ 1) = A + ~B = A + ~B + 1 - 1 = A + (~B + 1) - 1 = A - B - 1 ,

This means that the normal Add/Sub circuit can achieve the operation you require, all you need to do is change the input tied to '0' your Cin.

You could also design the desired circuit by using the usual method of writing all your possible inputs into a truth table, then deriving the Boolean equations for them, then simplifying them to get your final circuit equation.

If you are constrained to use decoders and NOR gates, the logic is a whole lot simpler if you redefine the carry inputs and outputs to be active-low instead of active-high. I'll present a solution based on this, and maybe it will provide some insight to your original problem.

^{simulate this circuit – Schematic created using CircuitLab}

Actually, it only requires one additional gate to make them active-high, but I'll leave that as an exercise for the reader.